Problem: Evaluate $\int\tan x\,dx\,$. Choose 1 answer: Choose 1 answer: (Choice A) A $\ln|\sec x+\tan x|+C$ (Choice B) B $\ln|\csc x|+C$ (Choice C) C $\ln|\sec x|+C$ (Choice D) D $\ln|\cos x|+C$
Answer: In this problem we will use the identity $~\tan x=\dfrac{\sin x}{\cos x}\,$. $\int\tan x\, dx =\int \dfrac{\sin x}{\cos x}\,dx$ Since the numerator is (essentially) the derivative of the denominator $-$ we're missing a minus-sign $-$ we can do the antidifferentiation without resorting to a $u$ -substitution. $\begin{aligned} \int\tan x\, dx &=\int \dfrac{\sin x}{\cos x}\,dx \\\\ &=-\int\dfrac{-\sin x}{\cos x}\,dx \\\\ &=-\ln|\cos x|+C \\\\ &=\ln|\cos x|^{-1}+C \\\\ &=\ln|\sec x|+C \end{aligned}$